Numerical Algebraic Geometry Boot Camp

Numerical Algebraic Geometry Boot Camp

!Dan Bates

Colorado State University !!!!

University of Notre Dame Workshop on…Numerical Algebraic Geometry May 23, 2016

t

t=0 t=1

• Introduce many of the basic

- structures,

- methods, and

- assumptions

of numerical algebraic geometry (and not the nitty-gritty theory underlying it)

!

• Show some Bertini Classic I/O

!

• Give us a common language for the next 2.5 days

!

Goals for this boot camp

1. Polynomial systems and their solution sets

2. Finding isolated solutions (homotopy continuation)

3. Advanced topics for isolated solutions

4. Finding positive-dimensional solution sets (briefly)

Game plan


A. Examples

B. Intuition from linear algebra

C. Some words

D. Bertini’s theorem

Game plan

Example 1: !Two solutions: !!! !!!

x2 � 1 = 0

5

x2 � 1 = 0

x = ±1

5

Polynomial systems and their solution sets

1-1 0

x4 � 1 = 0

x = ±i

x 2 R

5

A. Examples


C. Some words


Example 2: !Three solutions (1 real, 2 non-real) !

x3 � 1 = 0

x = ±i

5

1-1

i

-i

x4 � 1 = 0

x = ±i

x 2 C

5

Polynomial systems and their solution setsA. Examples


C. Some words


Example 3: !Four solutions (2 real, 2 non-real) !

x4 � 1 = 0

x = ±i

5

1-1

i

-i

x4 � 1 = 0

x = ±i

x 2 C

5



C. Some words


Example 4: !!Four solutions: !

x4 � 1 = 0

x = ±i

x 2 C

f(x, y) =

x2 � 1y2 � 4

�

5

1-1

2

-2

x4 � 1 = 0

x = ±i

x 2 C

f(x, y) =

x2 � 1y2 � 4

�

5

x4 � 1 = 0

x = ±i

y 2 C

f(x, y) =

x2 � 1y2 � 4

�

5

x4 � 1 = 0

x = ±i

y 2 C

f(x, y) =

x2 � 1y2 � 4

�

(±1,±2)

5



C. Some words


Example 5: !!Four solutions: !!! Visualization harder…

x4 � 1 = 0

x = ±i

y 2 C

g(x, y) =

x2 + 1y2 + 4

�

(±1,±2)

5

(1, 1), (�1,�1), (2,�1), (�2, 1)

f(x, y) =

f1 + f2 + f3f1 + 2f2 + f3

�

J =

2

64

@f1

@x1. . . @f1

@x

N

.

.

.

.

.

.

.

.

.

@f

N

@x1. . . @f

N

@x

N

3

75

(±i,±2i) 2 C2

6



C. Some words


Example 6:

x4 � 1 = 0

x = ±i

y 2 C

h(x, y) =

x2 + y2 � 1

x2 + (y + c)2 � 1

�

c = 1

c = 2

c = 3

5

x4 � 1 = 0

x = ±i

y 2 C

h(x, y) =

x2 + y2 � 1

x2 + (y + c)2 � 1

�

c = 1

c = 2

c = 3

5

x4 � 1 = 0

x = ±i

y 2 C

h(x, y) =

x2 + y2 � 1

x2 + (y + c)2 � 1

�

c = 1

c = 2

c = 3

5

x4 � 1 = 0

x = ±i

y 2 C

h(x, y) =

x2 + y2 � 1

x2 + (y + c)2 � 1

�

c = 1

c = 2

c = 3

5

x 2

2 complex



C. Some words


Example 7: !!Solutions: (a line) and (a point). ! (Two sets, two dimensions!) !

f =

x(x� 1)x(y � 1)

�

2

f =

x(x� 1)x(y � 1)

�

x = 0

x = y = 1

2

x4 � 1 = 0

x = ±i

y 2 C

g(x, y) =

x2 + 1y2 + 4

�

(1, 1)

5



C. Some words


Example 8: !!!!Solutions: ! Dimension 2: One surface Dimension 1: Three lines and one cubic curve Dimension 0: One point

“main”2013/3/17page 159

✐

✐

✐

✐

✐

✐

✐

✐

DRAFT

Chapter 8

Positive-DimensionalComponents

In Part II of the book, we return to the fundamental problem of “solving” a poly-nomial system,

f(z) :=

⎡

⎢⎣f1(z1, . . . , zN)

...fn(z1, . . . , zN)

⎤

⎥⎦ = 0. (8.1)

As opposed to the case of systems of linear equations, where a nonempty so-lution set will consist of exactly one linear space, the solution set of a polynomialsystem can have several separate components and these may have different dimen-sions23 (points, curves, surfaces, etc). Thus, the problem of “solving” a polynomialsystem includes the detection of all such components no matter what dimensioneach has. Solution components of dimension greater than zero (everything exceptisolated points) are said to be “positive-dimensional.”

For example, consider the following system of three polynomials in the vari-ables x, y, z:

f =

⎡

⎣(y − x2)(x2 + y2 + z2 − 1)(x− 2)(z − x3)(x2 + y2 + z2 − 1)(y − 2)

(z − x3)(y − x2)(x2 + y2 + z2 − 1)(z − 2)

⎤

⎦ . (8.2)

The solution set of three linear equations in three variables must be one point, oneline, or one plane. The complex solution set of (8.2) consists of a surface, fourcurves, and a point. If we were to apply one of the homotopies presented in Part Ito solve this system, we would be assured to find the isolated solution, (2, 2, 2), butthere would be no guarantee of finding any points on the other components.

In brief, Part II presents how numerical algebraic geometry treats full solutionsets, including their positive-dimensional components. This chapter introduces thebasic concepts and shows results obtained with Bertini on some simple examples.The next two chapters, Chapter 9 and Chapter 10, which go into details of the

23We gave a definition of the dimension of a component in §1.2.2.

159

(“Illustrative example” from Sommese-Verschelde-Wampler, SIAM J. Numer. An. ’01)



C. Some words


Linear systems vs. polynomial systems

Linear Polynomial

# solutions 0 or 1 0 or more

# dimensions 1 1 or more

solution components

point, line, plane, etc.

points, curves, surfaces, etc.

vanishing set for each equation

hyperplane hypersurface



C. Some words


Some words

Solution set: The set of all solutions of a polynomial system . Also called an algebraic set and sometimes a variety. Sometimes denoted . !Irreducible component: One “piece” of an algebraic set, e.g., a curve or a point. !Irreducible decomposition: , where ! is the dimension of , cycles through possible dimensions of irreducible components, is an index within dimension i, and the are the irreducible components.

f : CN ! CN

f1, . . . , fn 2 C[z1, . . . , zN ]

bz 2 CN

fi(bz) = 0 8i

f = (f1, . . . , fn)

R

N = n

N 6= n

⇡

Z = V(f) =D[

i=0

Zi =D[

i=0

[

j2⇤i

Zi,j

1

f : CN ! CN

f1, . . . , fn 2 C[z1, . . . , zN ]

bz 2 CN

fi(bz) = 0 8i

f = (f1, . . . , fn)

R

N = n

N 6= n

⇡

Z = V(f) =D[

i=0

Zi =D[

i=0

[

j2⇤i

Zi,j

1

f : CN ! CN

f1, . . . , fn 2 C[z1, . . . , zN ]

bz 2 CN

fi(bz) = 0 8i

f = (f1, . . . , fn)

R

N = n

N 6= n

⇡

Z = V(f) =D[

i=0

Zi =D[

i=0

[

j2⇤i

Zi,j

1

f : CN ! CN

f1, . . . , fn 2 C[z1, . . . , zN ]

bz 2 CN

fi(bz) = 0 8i

f = (f1, . . . , fn)

R

N = n

N 6= n

⇡

Z = V(f) =D[

i=0

Zi =D[

i=0

[

j2⇤i

Zi,j

1

f : CN ! CN

f1, . . . , fn 2 C[z1, . . . , zN ]

bz 2 CN

fi(bz) = 0 8i

f = (f1, . . . , fn)

R

N = n

N 6= n

⇡

Z = V(f) =D[

i=0

Zi =D[

i=0

[

j2⇤i

Zi,j

1

f : CN ! CN

f1, . . . , fn 2 C[z1, . . . , zN ]

bz 2 CN

fi(bz) = 0 8i

f = (f1, . . . , fn)

R

N = n

N 6= n

⇡

Z = V(f) =D[

i=0

Zi =D[

i=0

[

j2⇤i

Zi,j

1

x4 � 1 = 0

x = ±i

y 2 C

h(x, y) =

x2 + y2 � 1

x2 + (y + c)2 � 1

�

c = 1

c = 2

c = 3

V(f)

5

x4 � 1 = 0

x = ±i

y 2 C

h(x, y) =

x2 + y2 � 1

x2 + (y + c)2 � 1

�

c = 1

c = 2

c = 3

f

5



C. Some words



“main”2013/3/17page 159

✐

✐

✐

✐

✐

✐

✐

✐

DRAFT

Chapter 8



f(z) :=

⎡

⎢⎣f1(z1, . . . , zN)

...fn(z1, . . . , zN)

⎤

⎥⎦ = 0. (8.1)



f =

⎡

⎣(y − x2)(x2 + y2 + z2 − 1)(x− 2)(z − x3)(x2 + y2 + z2 − 1)(y − 2)

(z − x3)(y − x2)(x2 + y2 + z2 − 1)(z − 2)

⎤

⎦ . (8.2)




159

x4 � 1 = 0

x = ±i

y 2 C

h(x, y) =

x2 + y2 � 1

x2 + (y + c)2 � 1

�

c = 1

c = 2

c = 3

f

Z = Z0,0 [ Z1,0 [ Z1,1 [ Z1,2 [ Z1,3 [ Z2,0

5



C. Some words


Some more words

Regular/nonsingular solution: A solution of a polynomial system at which the Jacobian matrix (below) has full rank. Other solutions are called singular solutions. !!

x4 � 1 = 0

x = ±i

y 2 C

h(x, y) =

x2 + y2 � 1

x2 + (y + c)2 � 1

�

c = 1

c = 2

c = 3

f

5

(1, 1), (�1,�1), (2,�1), (�2, 1)

f(x, y) =

f1 + f2 + f3f1 + 2f2 + f3

�

J =

2

64

@f1

@x1. . . @f1

@x

N

.

.

.

.

.

.

.

.

.

@f

N

@x1. . . @f

N

@x

N

3

75

6

Example 6: !! (2 solutions): (full rank) ! (1 double solution): (rank-deficient)

x4 � 1 = 0

x = ±i

y 2 C

h(x, y) =

x2 + y2 � 1

x2 + (y + c)2 � 1

�

c = 1

c = 2

c = 3

5

f

Z = Z0,0 [ Z1,0 [ Z1,1 [ Z1,2 [ Z1,3 [ Z2,0

(1, 1), (�1,�1), (2,�1), (�2, 1)

f(x, y) =

f1 + f2 + f3f1 + 2f2 + f3

�

J =

2

64

@f1

@x1. . . @f1

@x

N

.

.

.

.

.

.

.

.

.

@f

N

@x1. . . @f

N

@x

N

3

75

(±i,±2i) 2 C2

i

L

J =

2x 2y2x 2y + 2c

�

6

x4 � 1 = 0

x = ±i

y 2 C

h(x, y) =

x2 + y2 � 1

x2 + (y + c)2 � 1

�

c = 1

c = 2

c = 3

5

x4 � 1 = 0

x = ±i

y 2 C

h(x, y) =

x2 + y2 � 1

x2 + (y + c)2 � 1

�

c = 1

c = 2

c = 3

5

f

Z = Z0,0 [ Z1,0 [ Z1,1 [ Z1,2 [ Z1,3 [ Z2,0

(1, 1), (�1,�1), (2,�1), (�2, 1)

f(x, y) =

f1 + f2 + f3f1 + 2f2 + f3

�

J =

2

64

@f1

@x1. . . @f1

@x

N

.

.

.

.

.

.

.

.

.

@f

N

@x1. . . @f

N

@x

N

3

75

(±i,±2i) 2 C2

i

L

J =

2x 2y2x 2y + 2c

�

J(0.866,�0.5) =

1.732 �11.732 1

�

6

f

Z = Z0,0 [ Z1,0 [ Z1,1 [ Z1,2 [ Z1,3 [ Z2,0

(1, 1), (�1,�1), (2,�1), (�2, 1)

f(x, y) =

f1 + f2 + f3f1 + 2f2 + f3

�

J =

2

64

@f1

@x1. . . @f1

@x

N

.

.

.

.

.

.

.

.

.

@f

N

@x1. . . @f

N

@x

N

3

75

(±i,±2i) 2 C2

i

L

J =

2x 2y2x 2y + 2c

�

J(0,�1) =

0 �20 2

�

6



C. Some words


Bertini’s Theorem

Polynomial systems can have different numbers of function (n) and variables (N). !If N > n (more variables), there will be no isolated solutions (handled later). !If N < n (more equations), we can replace with N random linear combinations of the polynomials of . Call the new system . Each solution of is a solution of , but might have more solutions (Bertini junk). !Example 9: !!!!!Solutions:

N = n

N 6= n

⇡

Z = V(f) =D[

i=0

Zi =D[

i=0

[

j2⇤i

Zi,j

f =

x(x� 1)x(y � 1)

�

x = 0

x = y = 1

z

||z � bz|| < FinalTol

N > n

2

N = n

N 6= n

⇡

Z = V(f) =D[

i=0

Zi =D[

i=0

[

j2⇤i

Zi,j

f =

x(x� 1)x(y � 1)

�

x = 0

x = y = 1

z


N > n

2

x4 � 1 = 0

x = ±i

y 2 C

h(x, y) =

x2 + y2 � 1

x2 + (y + c)2 � 1

�

c = 1

c = 2

c = 3

f

Z = Z0,0 [ Z1,0 [ Z1,1 [ Z1,2 [ Z1,3 [ Z2,0

5

N = n

N 6= n

⇡

Z = V(f) =D[

i=0

Zi =D[

i=0

[

j2⇤i

Zi,j

f =

x(x� 1)x(y � 1)

�

x = 0

x = y = 1

z


N > n

2

x4 � 1 = 0

x = ±i

y 2 C

h(x, y) =

x2 + y2 � 1

x2 + (y + c)2 � 1

�

c = 1

c = 2

c = 3

f

Z = Z0,0 [ Z1,0 [ Z1,1 [ Z1,2 [ Z1,3 [ Z2,0

5

x4 � 1 = 0

x = ±i

y 2 C

h(x, y) =

x2 + y2 � 1

x2 + (y + c)2 � 1

�

c = 1

c = 2

c = 3

f

Z = Z0,0 [ Z1,0 [ Z1,1 [ Z1,2 [ Z1,3 [ Z2,0

5

x4 � 1 = 0

x = ±i

y 2 C

h(x, y) =

x2 + y2 � 1

x2 + (y + c)2 � 1

�

f(x, y) =

2

4x2 � 1y2 � 1xy � 1

3

5

c = 1

c = 2

c = 3

f

Z = Z0,0 [ Z1,0 [ Z1,1 [ Z1,2 [ Z1,3 [ Z2,0

5

(1, 1), (�1,�1)

6

(1, 1), (�1,�1)

f(x, y) =

f1 + f2 + f3f1 + 2f2 + f3

�

6

(1, 1), (�1,�1), (2,�1), (�2, 1)

f(x, y) =

f1 + f2 + f3f1 + 2f2 + f3

�

6

❌ ❌☑ ☑

(not very random)



C. Some words






Game plan

!


A. Homotopy continuation in a nutshell

B. Start systems

C. Bells & whistles

D. Endgames

E. Bertini Classic (1.x)

!

Game plan

Finding isolated solutions

!!Given , we want to find all s.t. . !For each isolated solution, , we aim to compute a numerical approximation such that . !How do we accomplish this? Homotopy continuation.

f : CN ! CN

f1, . . . , fn 2 C[z1, . . . , zN ]

1

f : CN ! CN

f1, . . . , fn 2 C[z1, . . . , zN ]

bz 2 CN

fi(bz) = 0 8i

1

f : CN ! CN

f1, . . . , fn 2 C[z1, . . . , zN ]

bz 2 CN

fi(bz) = 0 8i

1

f : CN ! CN

f1, . . . , fn 2 C[z1, . . . , zN ]

bz 2 CN

fi(bz) = 0 8i

f = (f1, . . . , fn)

R

N = n

N 6= n

⇡

Z = V(f) =D[

i=0

Zi =D[

i=0

[

j2⇤i

Zi,j

1

f =

x(x� 1)x(y � 1)

�

x = 0

x = y = 1

z

2

f =

x(x� 1)x(y � 1)

�

x = 0

x = y = 1

z


2

A. Homotopy continuation B. Start systems C. Bells & whistles D. Endgames E. Bertini Classic (1.x)


Given polynomial system (the target system) homotopy continuation is a 3-step process: !1. Choose and solve a polynomial system (the start system)

based on characteristics of but relatively easy to solve. !2. Form the homotopy given by !! so that and . !3. Use numerical predictor-corrector methods to follow the solutions as marches from 1 to 0, one solution at a time. (Notice the parallelizability!) !!

f : CN ! CN

1

f : CN ! CN

g : CN ! CN

f1, . . . , fn 2 C[z1, . . . , zN ]

bz 2 CN

fi(bz) = 0 8i

f = (f1, . . . , fn)

R

N = n

N 6= n

⇡

1

f : CN ! CN

g : CN ! CN

H : CN ⇥ C ! CN

f1, . . . , fn 2 C[z1, . . . , zN ]

bz 2 CN

fi(bz) = 0 8i

f = (f1, . . . , fn)

R

N = n

N 6= n

⇡

1

f : CN ! CN

g : CN ! CN

H : CN ⇥ C ! CN

H(z, t) = f(z) · (1� t) + g(z) · t

H(z, 1) = g(z)

H(z, 0) = f(z)

f1, . . . , fn 2 C[z1, . . . , zN ]

bz 2 CN

fi(bz) = 0 8i

f = (f1, . . . , fn)

R

1

f : CN ! CN

g : CN ! CN

H : CN ⇥ C ! CN

H(z, t) = f(z) · (1� t) + g(z) · t

H(z, 1) = g(z)

H(z, 0) = f(z)

f1, . . . , fn 2 C[z1, . . . , zN ]

bz 2 CN

fi(bz) = 0 8i

f = (f1, . . . , fn)

R

1

f : CN ! CN

g : CN ! CN

H : CN ⇥ C ! CN

H(z, t) = f(z) · (1� t) + g(z) · t

H(z, 1) = g(z)

H(z, 0) = f(z)

f1, . . . , fn 2 C[z1, . . . , zN ]

bz 2 CN

fi(bz) = 0 8i

f = (f1, . . . , fn)

R

1

f : CN ! CN

g : CN ! CN

H : CN ⇥ C ! CN

H(z, t) = f(z) · (1� t) + g(z) · t

H(z, 1) = g(z)

H(z, 0) = f(z)

f1, . . . , fn 2 C[z1, . . . , zN ]

bz 2 CN

fi(bz) = 0 8i

f = (f1, . . . , fn)

R

1

f : CN ! CN

g : CN ! CN

H : CN ⇥ C ! CN

H(z, t) = f(z) · (1� t) + g(z) · t

H(z, 1) = g(z)

H(z, 0) = f(z)

f1, . . . , fn 2 C[z1, . . . , zN ]

bz 2 CN

fi(bz) = 0 8i

f = (f1, . . . , fn)

R

1



t

t=0 t=1

t 2 C

z 2 CN

1


t=1

t=0.75

t=0.5

t=0.25t=0

t=0.75

t=0.5

t=0.25

t=0

t=0.5

t=0.25

t=0

t=0.25

t=0

TROUBLE!


t

t=0 t=1

t 2 C

z 2 CN

1



Bézout: For a system with N polynomials and variables, the number of finite, !isolated solutions . (Over projective space, one can make !more exact statements….) !One choice of start system is the total degree or Bézout start system:

! (Recall the first 3 examples!) ! . This has exactly isolated, nonsingular, finite solutions. !This could be overkill — there might be many fewer solutions of ! !Different root counts lead to different start systems.

N > n

N < n

PN

SecurityMaxNorm

⇧Ni=1 deg(fi)

di = deg(fi)

g =

2

64zd11 � 1

.

.

.

zdNN � 1

3

75

3

Root counts and start systems

NY

i=1

deg(fi

)

di

= deg(fi

)

g =

2

64zd11 � 1

.

.

.

zdNN

� 1

3

75

f(x, y) =

xy � 1x2 � 1

�

A 2 CN⇥N

A = U⌃V ⇤

(A) = s

max

s

min

Ax = b

ACC ⇡ PREC� log10((A))

K : CN ! Cd

4

NY

i=1

deg(fi

)

di

= deg(fi

)

g =

2

64zd11 � 1

.

.

.

zdNN

� 1

3

75

f(x, y) =

xy � 1x2 � 1

�

A 2 CN⇥N

A = U⌃V ⇤

(A) = s

max

s

min

Ax = b


K : CN ! Cd

4

f : CN ! CN

g : CN ! CN

H : CN ⇥ C ! CN

H(z, t) = f(z) · (1� t) + �g(z) · t

� 2 C

H(z, 1) = g(z)

H(z, 0) = f(z)

f1, . . . , fN 2 C[z1, . . . , zN ]

q1, . . . qM 2 Ck

M >> 0

z 2 CN

1




Bézout (multihomogeneous version): The number of finite, isolated solutions is also bounded above by some combinatorial formula built from the multidegrees of the polynomials, when the variables are broken into multiple groups. !Depending on choice of variable groups, you might get fewer or more startpoints (typically more). !Example: !!The total degree is 4.

N > n

N < n

PN

SecurityMaxNorm

⇧Ni=1 deg(fi)

di = deg(fi)

g =

2

64zd11 � 1

.

.

.

zdNN � 1

3

75

g =

xy � 1x2 � 1

�

3




Example: !!The total degree is 4. !!!!!!The 2-homogeneous (or 2-hom) degree is 2, so we can build a start system with 2 nonsingular, isolated solutions and follow only 2 paths to find all solutions!

N > n

N < n

PN

SecurityMaxNorm

⇧Ni=1 deg(fi)

di = deg(fi)

g =

2

64zd11 � 1

.

.

.

zdNN � 1

3

75

g =

xy � 1x2 � 1

�

3

“main”2013/3/17page 81

✐

✐

✐

✐

✐

✐

✐

✐

DRAFT

5.1. Multihomogeneous homotopy 81

N 1 2 3 4 5 6 7 8Bell(N) 1 2 5 15 52 203 877 4140

N 9 10 11 12Bell(N) 21,147 115,975 678,570 4,213,597

N 25Bell(N) 4,638,590,332,229,999,353

Table 5.1. Illustration of the growth of Bell(N) as N grows.

example, the cubic x2z+xyz+1 is quadratic in (x, y) and linear in z, so a partitionof the variables as {(x, y), (z)} could be advantageous. If all three polynomials ina system of three polynomials in C[x, y, z] had this same degree structure, then itwould certainly be an advantage to partition the variables in this manner. Thesituation becomes more complicated when a subdivision reduces degrees in somepolynomials of the system but increases them in others.

5.1.1 Multihomogeneous Bezout number

The number of paths in a multihomogeneous homotopy, also called the multiho-mogeneous Bezout number17 of the system, is closely tied to multihomogeneousprojective spaces already mentioned in §4.5. To decide which grouping of the vari-ables to use, one may easily compute the multihomogeneous Bezout number of eachgrouping under consideration, and pick the smallest one. As already mentioned, thetotal number of possible groupings grows quickly with the number of variables, soBertini does not automate an exhaustive search. But with a little practice, a usercan identify and check the most promising candidates.

Instead of introducing a detailed notation for expressing the multihomoge-neous Bezout number, let us demonstrate the idea through examples. Let’s startwith the system

xy − 1 = 0, x2 − 1 = 0. (5.1)

There are two possible groupings, {(x, y)} and {(x), (y)}, each with its own Bezoutnumber. For each grouping, we make a multidegree table with one column pergroup and one row per equation, as follows. For the grouping {(x, y)}, the table is:

(x, y)

xy − 1 2x2 − 1 2

17For brevity, we may sometimes call this variously the ‘Bezout number’, the ‘Bezout count’, oreven simply the ‘root count’.

“main”2013/3/17page 82

✐

✐

✐

✐

✐

✐

✐

✐

DRAFT

82 Chapter 5. Types of Homotopies

since both equations have degree 2, while for {(x), (y)}, we have:

(x) (y)

xy − 1 1 1x2 − 1 2 0

indicating the degrees of the polynomials with respect to each variable group. Inthe latter case, we may say the polynomial xy − 1 has multidegree (1, 1), or moresimply, the polynomial is type (1, 1), while x2 − 1 is type (2, 0).

The rules for computing the Bezout number for a given degree table are asfollows. As a preparatory step, replace the variable groups with the dimension ofthe subspace they live in. That is, (x, y) ∈ C2, so the first table becomes

2

xy − 1 2x2 − 1 2

while since x ∈ C and y ∈ C, the second one becomes

1 1

xy − 1 1 1x2 − 1 2 0

Before evaluating the multihomogeneous Bezout count for these and otherexamples, let us describe the general methodology. First, lay out the multidegreetable for an m-homogeneous system as:

k1 · · · km

f1 d1,1 · · · d1,m...

......

fN dN,1 · · · dN,m

So kj is the dimension of the jth group of variables (Ckj or Pkj , as appropriate), anddi,j is the degree of the ith polynomial, fi, with respect to the jth variable group.With this notation, the rules for evaluating the m-homogeneous Bezout count are:

1. find all ways of choosing N of the di,j such that there is one chosen in eachrow and kj chosen in each column j;

2. for each such choice, multiply the chosen degrees; and

3. sum these products over all the choices.

Let us return to the example above. In the 1-homogeneous case of {(x, y)},there is only one combination of the multidegrees that is possible, so the Bezout




It is tempting to just try all possible variable groupings…DON’T! !The number of such groupings for n variables grows as the Bell number: !

“main”2013/3/17page 81

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✐

✐

✐

✐

DRAFT

5.1. Multihomogeneous homotopy 81

N 1 2 3 4 5 6 7 8Bell(N) 1 2 5 15 52 203 877 4140

N 9 10 11 12Bell(N) 21,147 115,975 678,570 4,213,597

N 25Bell(N) 4,638,590,332,229,999,353

Table 5.1. Illustration of the growth of Bell(N) as N grows.

example, the cubic x2z+xyz+1 is quadratic in (x, y) and linear in z, so a partitionof the variables as {(x, y), (z)} could be advantageous. If all three polynomials ina system of three polynomials in C[x, y, z] had this same degree structure, then itwould certainly be an advantage to partition the variables in this manner. Thesituation becomes more complicated when a subdivision reduces degrees in somepolynomials of the system but increases them in others.

5.1.1 Multihomogeneous Bezout number

The number of paths in a multihomogeneous homotopy, also called the multiho-mogeneous Bezout number17 of the system, is closely tied to multihomogeneousprojective spaces already mentioned in §4.5. To decide which grouping of the vari-ables to use, one may easily compute the multihomogeneous Bezout number of eachgrouping under consideration, and pick the smallest one. As already mentioned, thetotal number of possible groupings grows quickly with the number of variables, soBertini does not automate an exhaustive search. But with a little practice, a usercan identify and check the most promising candidates.

Instead of introducing a detailed notation for expressing the multihomoge-neous Bezout number, let us demonstrate the idea through examples. Let’s startwith the system

xy − 1 = 0, x2 − 1 = 0. (5.1)

There are two possible groupings, {(x, y)} and {(x), (y)}, each with its own Bezoutnumber. For each grouping, we make a multidegree table with one column pergroup and one row per equation, as follows. For the grouping {(x, y)}, the table is:

(x, y)

xy − 1 2x2 − 1 2

17For brevity, we may sometimes call this variously the ‘Bezout number’, the ‘Bezout count’, oreven simply the ‘root count’.

(See Wampler, Numer. Math., ’93 for more on this.)




Warning: I am entirely ignoring polyhedral homotopies, which can be very efficient in terms of the number of paths to be tracked but is sometimes expensive in terms of precomputation.

December 29, 2004 11:35 WSPC/Book Trim Size for 9.75in x 6.5in main975x65

118 Numerical Solution of Systems of Polynomials Arising in Engineering and Science

8.1 A Hierarchy of Structures

Coefficient-Parameter

Polynomial

Products

Newton

Polytopes

Monomial Products

Linear Products

Multihomogeneous

Total Degree

!

!

!

!

!

!

!

easierstartsystem

❄

specificity(fewer paths)

✻

❥A❥B

Fig. 8.1 Classes of Product Structures. Below line A, start systems can be solved using onlyroutines for solving linear systems of equations. Above line B, special methods must be designedcase-by-case.

Figure 8.1 shows a hierarchy of classes of special structures that are useful inconstructing homotopies. Each structure in the diagram is a member of the classabove it; for example, a total degree structure is a particular kind of multihomo-geneous structure. (In particular, as we will shortly see, it is a one-homogeneousstructure.) As we ascend the hierarchy, each class of structures presents more andmore possibilities for matching a particular target system that we wish to solve. Asindicated on the right of the diagram, this means that we can select a more specialstructure, usually with the aim of reducing the number of solution paths to trackin the homotopy. The trade-off we face in this ascent is indicated by the downwardpointing arrow on the left of the diagram: the lower structures allow us to selectstart systems that are easier to solve. For some problems, the ascent up the dia-gram pays handsomely in path reduction and may turn an intractable problem intoa solvable one. On the other hand, it can happen that solving a start system for ahigher structure can consume more computer time than is saved in path reduction.Unfortunately, even just counting the number of roots of the start system can beexpensive, so it is a matter of experience to decide the most advantageous spot inthis hierarchy to solve a particular problem.

Two dashed lines appear in Figure 8.1 to demarcate significant differences in thestart systems of homotopies respecting the various special structures. Below Line A,

(from Sommese-Wampler ‘05 book)


Finding isolated solutions!!

!!!!“2. Form the homotopy given by !!!so that and .” !!!!!

f : CN ! CN

g : CN ! CN

H : CN ⇥ C ! CN

f1, . . . , fn 2 C[z1, . . . , zN ]

bz 2 CN

fi(bz) = 0 8i

f = (f1, . . . , fn)

R

N = n

N 6= n

⇡

1

f : CN ! CN

g : CN ! CN

H : CN ⇥ C ! CN

H(z, t) = f(z) · (1� t) + g(z) · t

H(z, 1) = g(z)

H(z, 0) = f(z)

f1, . . . , fn 2 C[z1, . . . , zN ]

bz 2 CN

fi(bz) = 0 8i

f = (f1, . . . , fn)

R

1

f : CN ! CN

g : CN ! CN

H : CN ⇥ C ! CN

H(z, t) = f(z) · (1� t) + g(z) · t

H(z, 1) = g(z)

H(z, 0) = f(z)

f1, . . . , fn 2 C[z1, . . . , zN ]

bz 2 CN

fi(bz) = 0 8i

f = (f1, . . . , fn)

R

1

f : CN ! CN

g : CN ! CN

H : CN ⇥ C ! CN

H(z, t) = f(z) · (1� t) + g(z) · t

H(z, 1) = g(z)

H(z, 0) = f(z)

f1, . . . , fn 2 C[z1, . . . , zN ]

bz 2 CN

fi(bz) = 0 8i

f = (f1, . . . , fn)

R

1

!In fact, we use the homotopy: !!!where is chosen at random. !!!

f : CN ! CN

g : CN ! CN

H : CN ⇥ C ! CN

H(z, t) = f(z) · (1� t) + �g(z) · t

� 2 C

H(z, 1) = g(z)

H(z, 0) = f(z)

f1, . . . , fn 2 C[z1, . . . , zN ]

bz 2 CN

fi(bz) = 0 8i

f = (f1, . . . , fn)

1

f : CN ! CN

g : CN ! CN

H : CN ⇥ C ! CN

H(z, t) = f(z) · (1� t) + �g(z) · t

� 2 C

H(z, 1) = g(z)

H(z, 0) = f(z)

f1, . . . , fn 2 C[z1, . . . , zN ]

bz 2 CN

fi(bz) = 0 8i

f = (f1, . . . , fn)

1

The gamma trick



Detecting divergence!!Input can be non-homogeneous (mixed degrees in each polynomial) or homogeneous (all terms in any one polynomial have the same degree). !Either way, Bertini will homogenize (if necessary) and work over a random patch of . Thus, paths of infinite length become paths of finite length. !Even so, paths diverging to infinity often have highly singular endpoints, so it is preferable to avoid them. In Bertini, we kill any path that exceeds the threshold after t reaches 0.1. (Good for speed!)

N > n

N < n

PN

3

N > n

N < n

PN

SecurityMaxNorm

3



Adaptive steplength

t0t1 t0t1 t1X

Δt Δt/2


For matrix , the singular value decomposition (SVD) of is a decomposition with various properties. !For our purposes, the key is that is diagonal with nonnegative real entries called the singular values of . !Using this, we can define the condition number of as: !!!Wilkinson: When solving linear system ,

N > n

N < n

PN

SecurityMaxNorm

⇧Ni=1 deg(fi)

di = deg(fi)

g =

2

64zd11 � 1

.

.

.

zdNN � 1

3

75

g =

xy � 1x2 � 1

�

A 2 CN⇥N

3

N > n

N < n

PN

SecurityMaxNorm

⇧Ni=1 deg(fi)

di = deg(fi)

g =

2

64zd11 � 1

.

.

.

zdNN � 1

3

75

g =

xy � 1x2 � 1

�

A 2 CN⇥N

A = U⌃V ⇤

3

N > n

N < n

PN

SecurityMaxNorm

⇧Ni=1 deg(fi)

di = deg(fi)

g =

2

64zd11 � 1

.

.

.

zdNN � 1

3

75

g =

xy � 1x2 � 1

�

A 2 CN⇥N

A = U⌃V ⇤

3

N > n

N < n

PN

SecurityMaxNorm

⇧Ni=1 deg(fi)

di = deg(fi)

g =

2

64zd11 � 1

.

.

.

zdNN � 1

3

75

g =

xy � 1x2 � 1

�

A 2 CN⇥N

3

(A) = smax

smin

4

N > n

N < n

PN

SecurityMaxNorm

⇧Ni=1 deg(fi)

di = deg(fi)

g =

2

64zd11 � 1

.

.

.

zdNN � 1

3

75

g =

xy � 1x2 � 1

�

A 2 CN⇥N

3

(A) = smax

smin

Ax = b

4

(A) = smax

smin

Ax = b


4


Adaptive precision

N > n

N < n

PN

SecurityMaxNorm

⇧Ni=1 deg(fi)

di = deg(fi)

g =

2

64zd11 � 1

.

.

.

zdNN � 1

3

75

g =

xy � 1x2 � 1

�

A 2 CN⇥N

3


!!So, when the condition number gets high, we can increase precision to salvage accuracy. (There are many details….) !This isn’t free!! !The key point is that zones of ill-conditioning can cause numerical trouble but AMP reduces the size of these zones significantly. Of course, there is a limit on PREC and failure-causing pathologies can be constructed. !Bertini will either get through these zones or report path failures (rather than report incorrect results).

(A) = smax

smin

Ax = b


4


Adaptive precision



Endgames

We cannot avoid singularities and ill-conditioning at . !Interesting systems often have singular solutions, but we cannot track into . We use endgames built from interpolation or complex analysis. !!!!!!

t = 0

7

t = 0

7



Endgames

We cannot avoid singularities and ill-conditioning at . !Interesting systems often have singular solutions, but we cannot track into . We use endgames built from interpolation or complex analysis. !Two main options: !1. Power series endgame !2. Cauchy endgame

t = 0

7

t = 0

7


See Morgan-Sommese-Wampler, Num. Math. ’92 for details.


Endgames


t = 0

7

t = 0

7



Endgames


t = 0

7

t = 0

7


See Morgan-Sommese-Wampler, Adv. in Appl. Math. ’92 for details.


Endgames


t = 0

7

t = 0

7



Endgames

t = 0

7

t = 0

7




BertiniBertini is free to download and use, and the source code is publicly available. It may be downloaded (source code or binary) at bertini.nd.edu. !2002: Project started 2006: Initial beta release, during IMA special year. 2006-2015: Various extensions, releases. 2015: Latest version (1.5.2). ~ 8 downloads/day ~ 300 citations ! Bates Brake Hauenstein Sommese Wampler Next up: Bertini 2.0 (b2) Large development team (you??), C++, Boost, Eigen, GPL3, modules, etc. More on that soon….



Bertini Classic I/O

x4 � 1 = 0

x = ±i

y 2 C

h(x, y) =

x2 + y2 � 1

x2 + (y + c)2 � 1

�

c = 1

c = 2

c = 3

5



Bertini I/O


(screen output)


Bertini I/O


(main_data file)





Game plan

!

!


A. Parameter homotopies

B. Certification

C. Regeneration

Game plan

Advanced topics for isolated solutions

Given N polynomials, in N variables and k parameters , suppose we want to solve at M points in parameter space, , for some M>>0. !Option A (naïve): Solve at each of the M points from scratch, using a standard homotopy (such as a total degree homotopy). !Let P = # paths required to solve at each point in parameter space, L = lower bound on # divergent paths for each run from scratch. !

# paths: MP# wasted: ML

f : CN ! CN

g : CN ! CN

H : CN ⇥ C ! CN

H(z, t) = f(z) · (1� t) + �g(z) · t

� 2 C

H(z, 1) = g(z)

H(z, 0) = f(z)

f1, . . . , fN 2 C[z1, . . . , zN ]

bz 2 CN

fi(bz) = 0 8i

f = (f1(z, q) . . . , fN(z, q))

1

f : CN ! CN

g : CN ! CN

H : CN ⇥ C ! CN

H(z, t) = f(z) · (1� t) + �g(z) · t

� 2 C

H(z, 1) = g(z)

H(z, 0) = f(z)

f1, . . . , fN 2 C[z1, . . . , zN ]

z 2 CN

fi(bz) = 0 8i

f = (f1(z, q) . . . , fN(z, q))

1

f : CN ! CN

g : CN ! CN

H : CN ⇥ C ! CN

H(z, t) = f(z) · (1� t) + �g(z) · t

� 2 C

H(z, 1) = g(z)

H(z, 0) = f(z)

f1, . . . , fN 2 C[z1, . . . , zN ]

z 2 CN

q 2 Ck

fi(bz) = 0 8i

1

f : CN ! CN

g : CN ! CN

H : CN ⇥ C ! CN

H(z, t) = f(z) · (1� t) + �g(z) · t

� 2 C

H(z, 1) = g(z)

H(z, 0) = f(z)

f1, . . . , fN 2 C[z1, . . . , zN ]

z 2 CN

f = 0

q 2 Ck

1

f : CN ! CN

g : CN ! CN

H : CN ⇥ C ! CN

H(z, t) = f(z) · (1� t) + �g(z) · t

� 2 C

H(z, 1) = g(z)

H(z, 0) = f(z)

f1, . . . , fN 2 C[z1, . . . , zN ]

q1, . . . qM 2 Ck

M >> 0

z 2 CN

1

Parameter homotopies

A. Parameter homotopies

B. Certification

C. Regeneration

Option B (parameter homotopy):

1. Solve for randomly chosen q’ .

2. Homotope from q’ to qi for i=1,…,M, following all finite solutions from the Step 1 solve at q’.

!# paths: P + M(P-L)

# wasted: P !

!!!!!! See Li-Sauer-Yorke, SIAM J. Num. An. ’89 and Morgan-Sommese, Appl. Math. and Comp. ’89.

f = 0

q0 2 Ck

fi(bz) = 0 8i

f = (f1(z, q0) . . . , fN(z, q0))

R

N = n

N 6= n

⇡

Z = V(f) =D[

i=0

Zi =D[

i=0

[

j2⇤i

Zi,j

f =

x(x� 1)x(y � 1)

�

2

f = 0

q0 2 Ck

fi(bz) = 0 8i

f = (f1(z, q0) . . . , fN(z, q0))

R

N = n

N 6= n

⇡

Z = V(f) =D[

i=0

Zi =D[

i=0

[

j2⇤i

Zi,j

f =

x(x� 1)x(y � 1)

�

2


Advanced topics for isolated solutionsA. Parameter homotopies

B. Certification

C. Regeneration

!!!

Example: P = 10, L = 5, M = 1,000,000: !!!!!!!Of course, if L=0 (not common), Option A is marginally better, at least in terms of # paths.

Option A # paths: MP = 10,000,000 # wasted: ML = 5,000,000

Option B # paths: P + M(P-L) = 5,000,010 # wasted: P = 10



B. Certification

C. Regeneration

Schematic !

qi !

q’ !



B. Certification

C. Regeneration

Schematic: Step 1 !

qi !

q’ !



B. Certification

C. Regeneration

Schematic: Step 2 !

qi !

q’ !

… !



B. Certification

C. Regeneration

Problem: Standard NAG software options include parameter homotopy capabilities, but: * parallelization is within the run (not between runs), * failure mitigation is not automated (a headache), and * each run includes startup costs, e.g., parsing. !Solution: Paramotopy D. Bates, D. Brake (Notre Dame), and M. Niemerg (IBM/Oak Ridge) www.paramotopy.com ! Free to use, source code publicly available C++, Boost Uses Bertini as a library To be incorporated into Bertini 2.x in some form.



B. Certification

C. Regeneration

Certification

!Given some approximation of a nonsingular isolated solution of a polynomial system, Newton’s method will: diverge, converge slowly, or converge quadratically. !!!!!

Basins of attraction for z5 + 1 = 0 from MIT Open Courseware (Intro to Matlab Programming)


B. Certification

C. Regeneration

Certification

!Given some approximation of a nonsingular isolated solution of a polynomial system, Newton’s method will: diverge, converge slowly, or converge quadratically. !The region of quadratic convergence is quite small. !It is possible to prove that an approximation is within this region of quadratic convergence for some solution. This is the crux of alpha theory, though the certifiable region is extremely small!


B. Certification

C. Regeneration

Certification

Two basic options (more introduced recently): ! 1. Certify as you go through a homotopy (Beltran-Leykin, NAG4M2) or 2. Post-certify (Hauenstein-Sottile, alphaCertified). !See the upcoming talk of Nick Hein & Alan Liddell.


B. Certification

C. Regeneration

Regeneration

This is the latest equation-by-equation solver. Replaces standard homotopy continuation with a number of simpler homotopies. !!!!!!!!!!!! See Hauenstein-Sommese-Wampler, Math. Comp. ’10 for more details.


B. Certification

C. Regeneration





Game plan

!

!

!


A. Slicing and the numerical irreducible decomposition

B. Bertini Classic I/O

C. Sampling

D. Real solutions

Game plan

Finding positive-dimensional solution sets!!!!!!Recall: , where: ! is the dimension of , cycles through possible dimensions of irreducible components, is an index within dimension i, and the are the irreducible components.

f : CN ! CN

f1, . . . , fn 2 C[z1, . . . , zN ]

bz 2 CN

fi(bz) = 0 8i

f = (f1, . . . , fn)

R

N = n

N 6= n

⇡

Z = V(f) =D[

i=0

Zi =D[

i=0

[

j2⇤i

Zi,j

1

f : CN ! CN

f1, . . . , fn 2 C[z1, . . . , zN ]

bz 2 CN

fi(bz) = 0 8i

f = (f1, . . . , fn)

R

N = n

N 6= n

⇡

Z = V(f) =D[

i=0

Zi =D[

i=0

[

j2⇤i

Zi,j

1

f : CN ! CN

f1, . . . , fn 2 C[z1, . . . , zN ]

bz 2 CN

fi(bz) = 0 8i

f = (f1, . . . , fn)

R

N = n

N 6= n

⇡

Z = V(f) =D[

i=0

Zi =D[

i=0

[

j2⇤i

Zi,j

1

f : CN ! CN

f1, . . . , fn 2 C[z1, . . . , zN ]

bz 2 CN

fi(bz) = 0 8i

f = (f1, . . . , fn)

R

N = n

N 6= n

⇡

Z = V(f) =D[

i=0

Zi =D[

i=0

[

j2⇤i

Zi,j

1

f : CN ! CN

f1, . . . , fn 2 C[z1, . . . , zN ]

bz 2 CN

fi(bz) = 0 8i

f = (f1, . . . , fn)

R

N = n

N 6= n

⇡

Z = V(f) =D[

i=0

Zi =D[

i=0

[

j2⇤i

Zi,j

1

f : CN ! CN

f1, . . . , fn 2 C[z1, . . . , zN ]

bz 2 CN

fi(bz) = 0 8i

f = (f1, . . . , fn)

R

N = n

N 6= n

⇡

Z = V(f) =D[

i=0

Zi =D[

i=0

[

j2⇤i

Zi,j

1

!!!!!!(This is the irreducible decomposition of .) !!

R

N = n

N 6= n

⇡

Z = V(f) =D[

i=0

Zi =D[

i=0

[

j2⇤i

Zi,j

f =

x(x� 1)x(y � 1)

�

x = 0

x = y = 1

z


2

!For each positive-dimensional irreducible component, , we aim to find numerical approximations to some number of generic points on .

f : CN ! CN

f1, . . . , fn 2 C[z1, . . . , zN ]

bz 2 CN

fi(bz) = 0 8i

f = (f1, . . . , fn)

R

N = n

N 6= n

⇡

Z = V(f) =D[

i=0

Zi =D[

i=0

[

j2⇤i

Zi,j

1

f : CN ! CN

f1, . . . , fn 2 C[z1, . . . , zN ]

bz 2 CN

fi(bz) = 0 8i

f = (f1, . . . , fn)

R

N = n

N 6= n

⇡

Z = V(f) =D[

i=0

Zi =D[

i=0

[

j2⇤i

Zi,j

1

A. Numerical irreducible decomposition

B. Bertini I/O C. Sampling D. Real solutions

Key fact: Given irreducible component of dimension , for almost every choice of linear space of codimension , intersects in a set of a particular number of points. That number is the degree of . !So, to find deg( ) points on , we can append linears to . We refer to this operation as slicing. !To find points on all components, we can just loop through for all . !!

f : CN ! CN

f1, . . . , fn 2 C[z1, . . . , zN ]

bz 2 CN

fi(bz) = 0 8i

f = (f1, . . . , fn)

R

N = n

N 6= n

⇡

Z = V(f) =D[

i=0

Zi =D[

i=0

[

j2⇤i

Zi,j

1

f : CN ! CN

f1, . . . , fn 2 C[z1, . . . , zN ]

bz 2 CN

fi(bz) = 0 8i

f = (f1, . . . , fn)

R

N = n

N 6= n

⇡

Z = V(f) =D[

i=0

Zi =D[

i=0

[

j2⇤i

Zi,j

1

f : CN ! CN

f1, . . . , fn 2 C[z1, . . . , zN ]

bz 2 CN

fi(bz) = 0 8i

f = (f1, . . . , fn)

R

N = n

N 6= n

⇡

Z = V(f) =D[

i=0

Zi =D[

i=0

[

j2⇤i

Zi,j

1

f : CN ! CN

f1, . . . , fn 2 C[z1, . . . , zN ]

bz 2 CN

fi(bz) = 0 8i

f = (f1, . . . , fn)

R

N = n

N 6= n

⇡

Z = V(f) =D[

i=0

Zi =D[

i=0

[

j2⇤i

Zi,j

1

R

N = n

N 6= n

⇡

Z = V(f) =D[

i=0

Zi =D[

i=0

[

j2⇤i

Zi,j

f =

x(x� 1)x(y � 1)

�

x = 0

x = y = 1

z


2

f : CN ! CN

f1, . . . , fn 2 C[z1, . . . , zN ]

bz 2 CN

fi(bz) = 0 8i

f = (f1, . . . , fn)

R

N = n

N 6= n

⇡

Z = V(f) =D[

i=0

Zi =D[

i=0

[

j2⇤i

Zi,j

1

(1, 1), (�1,�1), (2,�1), (�2, 1)

f(x, y) =

f1 + f2 + f3f1 + 2f2 + f3

�

J =

2

64

@f1

@x1. . . @f1

@x

N

.

.

.

.

.

.

.

.

.

@f

N

@x1. . . @f

N

@x

N

3

75

(±i,±2i) 2 C2

i

L

6

(1, 1), (�1,�1), (2,�1), (�2, 1)

f(x, y) =

f1 + f2 + f3f1 + 2f2 + f3

�

J =

2

64

@f1

@x1. . . @f1

@x

N

.

.

.

.

.

.

.

.

.

@f

N

@x1. . . @f

N

@x

N

3

75

(±i,±2i) 2 C2

i

L

6

(1, 1), (�1,�1), (2,�1), (�2, 1)

f(x, y) =

f1 + f2 + f3f1 + 2f2 + f3

�

J =

2

64

@f1

@x1. . . @f1

@x

N

.

.

.

.

.

.

.

.

.

@f

N

@x1. . . @f

N

@x

N

3

75

(±i,±2i) 2 C2

i

L

6

(1, 1), (�1,�1), (2,�1), (�2, 1)

f(x, y) =

f1 + f2 + f3f1 + 2f2 + f3

�

J =

2

64

@f1

@x1. . . @f1

@x

N

.

.

.

.

.

.

.

.

.

@f

N

@x1. . . @f

N

@x

N

3

75

(±i,±2i) 2 C2

i

L

6

(1, 1), (�1,�1), (2,�1), (�2, 1)

f(x, y) =

f1 + f2 + f3f1 + 2f2 + f3

�

J =

2

64

@f1

@x1. . . @f1

@x

N

.

.

.

.

.

.

.

.

.

@f

N

@x1. . . @f

N

@x

N

3

75

(±i,±2i) 2 C2

i

L

6

Finding positive-dimensional solution setsA. Numerical irreducible

decomposition B. Bertini I/O C. Sampling D. Real solutions

(1, 1), (�1,�1), (2,�1), (�2, 1)

f(x, y) =

f1 + f2 + f3f1 + 2f2 + f3

�

J =

2

64

@f1

@x1. . . @f1

@x

N

.

.

.

.

.

.

.

.

.

@f

N

@x1. . . @f

N

@x

N

3

75

(±i,±2i) 2 C2

i

L

6

Problem 1: We could pick up points on higher-dimensional components. Problem 2: We could find points on multiple i-dimensional components. !!!!!

Example: Suppose there are two curves and a surface. When we slice for the curves, we will find points on both curves and also on the surface. !Solution 1: Start at the top dimension and work your way down. Use a membership test on points in lower dimensions to see if they sit on the higher-dimensional components already found. !Solution 2: Carry out an equidimensional decomposition, using monodromy and the trace test. !!!



In fact, there is a clever way to string the homotopies together, called a cascade of homotopies. (There are more recent approaches, too.) !All told, the goal is to have deg witness points on each component , yielding witness point set !For each component, put the linear functions, the witness points, and the original functions together and you have a witness set for the component: !!Then, the numerical irreducible decomposition is the union of all such sets for all irreducible components: !!!Recall Example 8 again:

f : CN ! CN

f1, . . . , fn 2 C[z1, . . . , zN ]

bz 2 CN

fi(bz) = 0 8i

f = (f1, . . . , fn)

R

N = n

N 6= n

⇡

Z = V(f) =D[

i=0

Zi =D[

i=0

[

j2⇤i

Zi,j

1

f = 0

q0 2 Ck

fi

(bz) = 0 8i

f = (f1(z, q0) . . . , fN(z, q0))

R

N = n

N 6= n

⇡

W =D[

i=0

Wi

=D[

i=0

[

j2⇤i

Wi,j

Wi,j

= (f, Li,j

,Wi,j

)

2

Wi,j

= Zi,j

\ Li

f =

x(x� 1)x(y � 1)

�

x = 0

x = y = 1

z


N > n

N < n

PN

SecurityMaxNorm

3

f : CN ! CN

f1, . . . , fn 2 C[z1, . . . , zN ]

bz 2 CN

fi(bz) = 0 8i

f = (f1, . . . , fn)

R

N = n

N 6= n

⇡

Z = V(f) =D[

i=0

Zi =D[

i=0

[

j2⇤i

Zi,j

1



f = 0

q0 2 Ck

fi

(bz) = 0 8i

f = (f1(z, q0) . . . , fN(z, q0))

R

N = n

N 6= n

⇡

W =D[

i=0

Wi

=D[

i=0

[

j2⇤i

Wi,j

Wi,j

= (f, Li

,Wi,j

)

2


“main”2013/3/17page 159

✐

✐

✐

✐

✐

✐

✐

✐

DRAFT

Chapter 8



f(z) :=

⎡

⎢⎣f1(z1, . . . , zN)

...fn(z1, . . . , zN)

⎤

⎥⎦ = 0. (8.1)



f =

⎡

⎣(y − x2)(x2 + y2 + z2 − 1)(x− 2)(z − x3)(x2 + y2 + z2 − 1)(y − 2)

(z − x3)(y − x2)(x2 + y2 + z2 − 1)(z − 2)

⎤

⎦ . (8.2)




159



From Sommese-Verschelde-Wampler, SIAM J. Num. An. ’01.

Cascade of homotopies for computing the numerical irreducible decomposition of the illustrative example. !Notice that we find the degree number of points on each component.



Bertini Classic I/O



(screen output)

By following the solutions as we move the linears from a witness set, we pick up more points on the same irreducible component. This is called component sampling. !In this way, you can find many points on a curve (or other irreducible component) quite rapidly.

Sampling



For real applications, people typically want real solutions! !This is very difficult. There are several recent, limited techniques using homotopy methods: ! * Khovanskii-Rolle continuation for fewnomials (Bates-Sottile, KhRo) ! * Seidenberg-like methods (Hauenstein) ! * Real cellular decompositions (Lu-Bates-Sommese-Wampler, Besana-Di Rocco-Hauenstein-Sommese-Wampler, Bates-Brake-Hao-Hauenstein-Sommese-Wampler, BertiniReal) !There are non-homotopy numerical methods, too (cellular exclusion, cylindrical decomposition, etc.). !This is still a major open problem!

Real solutions



Here’s a sketch of how Bertini Real works: Curves first: !!!

Given

1. a witness set for complex curve Z and

2. randomly chosen projection � = Az (A random),

we want to find real points or curves inside Z.

The isolated real solutions and the critical points of any real curve will beisolated real solutions of

6



This last bit is a polynomial system that we can solve. Let’s switch to aschematic drawing, in R2 for simplicity:Notice that there are three real curves (two compact) and two points thatwe want to find.The polynomial system two slides back will find all inherent singular pointsand singular points of the projection (where topology changes occur).Projecting these down via �, we get points “downstairs” with fibers contain-ing singular points. Nothing “interesting” happens between these points.Choosing midpoints between these points “downstairs,” we may find allpoints in the fiber by moving the slices of the witness set for Z to the appro-priate linears.Moving these fiber points back and forth (moving the midpoints downstairsto the left and right to the black points), we get a schematic representation ofthe curve. This is impossible if the real curve is singular, without deflation.

7



This last bit is a polynomial system that can easily solve. Let’s switch to aschematic drawing, in R2 for simplicity:Notice that there are three real curves (two compact) and two points thatwe want to find.The polynomial system two slides back will find all inherent singular pointsand singular points of the projection (where topology changes occur).Projecting these down via �, we get points “downstairs” with fibers contain-ing singular points. Nothing “interesting” happens between these points.Choosing midpoints between these points “downstairs,” we may find allpoints in the fiber by moving the slices of the witness set for Z to the appro-priate linears.

6



This last bit is a polynomial system that we can solve. Let’s switch to aschematic drawing, in R2 for simplicity:Notice that there are three real curves (two compact) and two points thatwe want to find.The polynomial system three slides back will find all isolated real points andsingular points of the projection (where topology changes occur).Projecting these down via � (just function evaluation), we get points“downstairs” with fibers containing singular points. Nothing “interesting”happens between these points.Choosing midpoints between these points “downstairs,” we may find allpoints in the fiber by moving the slices of the witness set for Z to the appro-priate linears. There is a special (parameter) homotopy for this move.Moving these fiber points back and forth (moving the midpoints downstairsto the left and right to the black points), we play connect-the-dots and get aschematic representation of the curve. If the real curve is singular deflationis necessary.

9



This last bit is a polynomial system that we can solve. Let’s switch to aschematic drawing, in R2 for simplicity:Notice that there are three real curves (two compact) and two points thatwe want to find.The polynomial system two slides back will find all inherent singular pointsand singular points of the projection (where topology changes occur).Projecting these down via � (just function evaluation), we get points“downstairs” with fibers containing singular points. Nothing “interesting”happens between these points.Choosing midpoints between these points “downstairs,” we may find allpoints in the fiber by moving the slices of the witness set for Z to the appro-priate linears.Moving these fiber points back and forth (moving the midpoints downstairsto the left and right to the black points), we get a schematic representation ofthe curve. This is impossible if the real curve is singular, without deflation.

7



This last bit is a polynomial system that we can solve. Let’s switch to aschematic drawing, in R2 for simplicity:Notice that there are three real curves (two compact) and two points thatwe want to find.The polynomial system two slides back will find all inherent singular pointsand singular points of the projection (where topology changes occur).Projecting these down via � (just function evaluation), we get points“downstairs” with fibers containing singular points. Nothing “interesting”happens between these points.Choosing midpoints between these points “downstairs,” we may find allpoints in the fiber by moving the slices of the witness set for Z to the appro-priate linears. There is a special (parameter) homotopy for this move.Moving these fiber points back and forth (moving the midpoints downstairsto the left and right to the black points), we get a schematic representation ofthe curve. This is impossible if the real curve is singular, without deflation.

7



This last bit is a polynomial system that we can solve. Let’s switch to aschematic drawing, in R2 for simplicity:Notice that there are three real curves (two compact) and two points thatwe want to find.The polynomial system two slides back will find all inherent singular pointsand singular points of the projection (where topology changes occur).Projecting these down via � (just function evaluation), we get points“downstairs” with fibers containing singular points. Nothing “interesting”happens between these points.Choosing midpoints between these points “downstairs,” we may find allpoints in the fiber by moving the slices of the witness set for Z to the appro-priate linears. There is a special (parameter) homotopy for this move.Moving these fiber points back and forth (moving the midpoints downstairsto the left and right to the black points), we play connect-the-dots and get aschematic representation of the curve. If the real curve is singular deflationis necessary.

7



The surface case relies heavily on the curve case.

In the curve case, we find projections of critical points, then connect thedots in the fiber.

In the surface case, we find projections of critical points, constructcritical lines (giving critical curves), then connect the 2-D patches to thecritical curves.

Here’s a picture:

9

Now for the surface case: !!!



This surface has a hole anda pinchpoint.

Red dots on x-axis correspondto topology changes upstairs.

The blue curve on the floor isthe projection of the critical curve.

We use the curve algorithm (previousslides) to decompose the critical curve.

9



Here, the gray arcs lie abovethe critical slices in the floor of theprevious figure. Again, the curvealgorithm is used for this.

The purple arcs lie aboveundrawn midpoints betweencritical points.

The blue arcs show how themidpoints of the gray and purplearcs connect up. All light blue dotswere found via sampling.

10



The hardest part of thesurface case is playing“connect-the-edges.”

For this, we build ahomotopy that makescritical points stay on thecritical curve and movesthe 2D cell midpoint to theedge midpoints.

12

See Dan Brake’s talk for how to make use of this….



General references for numerical algebraic geometry ! Sommese-Wampler, Numerical solution of polynomial systems arising in science and engineering, World Scientific, 2005. !Bates-Hauenstein-Sommese-Wampler, Numerically solving polynomial systems with Bertini, SIAM, 2013. !More specific references: !Path-tracking: Allgower-Georg, SIAM, 2004. Polyhedral methods: Li, Acta Numerica, 2003. Algebraic kinematics: Sommese-Wampler, Acta Numerica, 2013. Many more — just ask!

Finding positive-dimensional solution sets

THANKS!!! (These slides are already posted on my website.)

!!!!!!!!!!!!!!!!

Partially supported by NSF ACI-1440467, FA8650-13-1-7317, and the IMA.

Numerical Algebraic Geometry Boot Camp

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